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Liono4ka [1.6K]

3 years ago

7

You observe three carts moving to the left. Cart A moves to the left at nearly constant speed. Cart B moves to the left, gradual

ly speeding up. Cart C moves to the left, gradually slowing down. Which cart or carts, if any, experience a net force to the left? Cart A Cart C Cart B

1 answer:

Lady bird [3.3K]3 years ago

6 0

Answer:cart B

Explanation:

For cart A speed is constant therefore there is no acceleration because acceleration is rate of change of velocity

thus there is no net force

For cart B there is change in velocity in the left direction , so there is net acceleration towards left

$Force=mass\times acceleration$

so there is net force in the left direction

For cart C there is decrease in velocity i.e. negative acceleration or deceleration . Therefore there is a net force towards right which opposes the motion

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Let surface S be the boundary of the solid object enclosed by x^2+z^2=4, x+y=6, x=0, y=0, and z=0. and, let f(x,y,z)=(3x)i+(x+y+

babunello [35]

a. I've attached a plot of the surface. Each face is parameterized by

• $\mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j$ with $0\le x\le2$ and $0\le y\le6-x$ ;

• $\mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ;

• $\mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k$ with $0\le y\le 6$ and $0\le z\le2$ ;

• $\mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ; and

• $\mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k$ with $0\le u\le\frac\pi2$ and $0\le y\le6-2\cos u$ .

b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.

$\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k$

$\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j$

$\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i$

$\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j$

$\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k$

Then integrate the dot product of <em>f</em> with each normal vector over the corresponding face.

$\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0$

$\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du$

$\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8$

$\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz$

$=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0$

$\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du$

$=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi$

$\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du$

$=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24$

c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.

Alternatively, since <em>S</em> is closed, we can find the total flux by applying the divergence theorem.

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV$

where <em>R</em> is the interior of <em>S</em>. We have

$\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7$

The integral is easily computed in cylindrical coordinates:

$\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2$

$\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3$

as expected.

4 0

3 years ago

9. What do people playing pool use to determine their shots?

jek_recluse [69]

Angles, they line up their pool que with the pocket and make the shot

7 0

3 years ago

Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea

Alexxx [7]

Answer:

acceleration are

hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

Em₀ = U = m g h

Final point. To get off the ramp

Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

Em₀ = Em_f

mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

v = w r

w = v / r

we substitute

mg h = ½ v² (m + I / r²)

v² = 2 gh $\frac{m}{m+ \frac{I}{r^2} }$

v² = 2gh $\frac{1}{1 + \frac{I}{m r^2} }$

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

v² = v₀² + 2 a L

where L is the length of the plane

v² = 2 a L

a = v² / 2L

we substitute

a = $g \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }$

let's use trigonometry

sin θ = h / L

we substitute

a = g sin θ \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

I = m r²

we look for the acerleracion

a₁ = g sin θ $\frac{1}{1 + \frac{mr^2 }{m r^2 } }$ 1/1 + mr² / mr² =

a₁ = g sin θ ½

b) solid cylinder

I = ½ m r²

a₂ = g sin θ $\frac{1}{1 + \frac{1}{2} \frac{mr^2}{mr^2} }$ = g sin θ $\frac{1}{1+ \frac{1}{2} }$

a₂ = g sin θ ⅔

c) hollow sphere

I = 2/3 m r²

a₃ = g sin θ $\frac{1}{1 + \frac{2}{3} }$

a₃ = g sin θ $\frac{3}{5}$

d) solid sphere

I = 2/5 m r²

a₄ = g sin θ $\frac{1 }{1 + \frac{2}{5} }$

a₄ = g sin θ $\frac{5}{7}$

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ $\frac{105}{210}$

b) a₂ = g sinθ ⅔ = g sin θ $\frac{140}{210}$

c) a₃ = g sin θ $\frac{3}{5}$ = g sin θ $\frac{126}{210}$

d) a₄ = g sin θ $\frac{5}{7}$ = g sin θ $\frac{150}{210}$

the order of acceleration from lower to higher is

a₁ <a₃ <a₂ <a₄

acceleration are

hollow cylinder < hollow sphere < solid cylinder < solid sphere

8 0

3 years ago

What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resti

Nataliya [291]

Archimedes principle states that

F1 / A1 = F2 / A2

F2 = (A2 / A1) * F1

Also, formula for the force is F = mg. Formula for the area of the cylinder is A = πr^2, therefore we get

F2 = (πr2^2 / πr1^2) * mg

Since the diameter of the cylinders are 2 cm and 24 cm, r1 = 12 and r2 = 1.

Substituting the values to the derived equation, we get

F2 = (π 1^2 / π 12^2) * 2400 * 9.8

F2 = 163.3333 N

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6 0

3 years ago

The unit used to measure electric current is the ampere (A). Now, assume that the current delivered at a wall socket reaches the

pav-90 [236]

Answer:

T = 0.017s

Explanation:

period is the time it takes a particle to make one oscillation

An electric current is periodic in nature

The current reaches 3.8A ten times.

So there must have been 10 cycles (10 periods) in 0.17s. let 'T' be the period:

$T=\frac{t}{n}$

t is the total time interval

n is the number of oscillations

$T=\frac{0.17}{10}$

10T = 0.17

T = 0.17/10 = 0.017s

8 0

3 years ago